NCERT Class 11 Mathematics — Statistics, Exercise 13.1. All 12 questions solved.

Mean deviation measures spread as the average distance from a central value:

$$\text{M.D.}(\bar x) = \frac{\sum f_i\left|x_i – \bar x\right|}{\mathrm N}, \qquad \text{M.D.}(\mathrm M) = \frac{\sum f_i\left|x_i – \mathrm M\right|}{\mathrm N}$$

The three data types need slightly different handling:

• Raw data: $\bar x = \dfrac{\sum x_i}{n}$; the median is the middle value of the sorted list (or the average of the two middle ones).
• Discrete frequency: the median is the $x_i$ at which the cumulative frequency first reaches $\tfrac{\mathrm N}{2}$.
• Grouped data: use class midpoints for $x_i$, and

$$\mathrm M = l + \frac{\frac{\mathrm N}{2} – \mathrm C}{f} \times h$$

where $l$ is the lower limit of the median class, $\mathrm C$ the cumulative frequency before it, $f$ its frequency and $h$ the class width.

Key insight. The modulus is what makes this work — without it the deviations from the mean would cancel to exactly zero, every time. So always take the absolute value before summing, never after. A sum of $|x_i – \bar x|$ that comes out as $0$ is a sure sign the moduli were dropped.

Find the mean deviation about the mean for the data in questions 1 and 2.

Question 1

$4,\ 7,\ 8,\ 9,\ 10,\ 12,\ 13,\ 17$

Solution. There are $8$ observations, summing to $80$:

$$\bar x = \frac{80}{8} = 10$$

Now the absolute deviations from $10$:

$$6,\ 3,\ 2,\ 1,\ 0,\ 2,\ 3,\ 7$$

Their sum is $24$, so

$$\text{M.D.}(\bar x) = \frac{24}{8} = 3$$

Question 2

$38,\ 70,\ 48,\ 40,\ 42,\ 55,\ 63,\ 46,\ 54,\ 44$

Solution. Ten observations summing to $500$:

$$\bar x = \frac{500}{10} = 50$$

Absolute deviations from $50$:

$$12,\ 20,\ 2,\ 10,\ 8,\ 5,\ 13,\ 4,\ 4,\ 6$$

Sum $= 84$, so

$$\text{M.D.}(\bar x) = \frac{84}{10} = 8.4$$

Find the mean deviation about the median for the data in questions 3 and 4.

Question 3

$13,\ 17,\ 16,\ 14,\ 11,\ 13,\ 10,\ 16,\ 11,\ 18,\ 12,\ 17$

Solution. Sort first — this is the step that cannot be skipped:

$$10,\ 11,\ 11,\ 12,\ 13,\ 13,\ 14,\ 16,\ 16,\ 17,\ 17,\ 18$$

With $n = 12$ (even), the median is the average of the $6$th and $7$th values:

$$\mathrm M = \frac{13 + 14}{2} = 13.5$$

Absolute deviations from $13.5$:

$$3.5,\ 2.5,\ 2.5,\ 1.5,\ 0.5,\ 0.5,\ 0.5,\ 2.5,\ 2.5,\ 3.5,\ 3.5,\ 4.5$$

Sum $= 28$, so

$$\text{M.D.}(\mathrm M) = \frac{28}{12} = \frac73 \approx 2.33$$

Question 4

$36,\ 72,\ 46,\ 42,\ 60,\ 45,\ 53,\ 46,\ 51,\ 49$

Solution. Sorted:

$$36,\ 42,\ 45,\ 46,\ 46,\ 49,\ 51,\ 53,\ 60,\ 72$$

With $n = 10$, the median is the average of the $5$th and $6$th:

$$\mathrm M = \frac{46 + 49}{2} = 47.5$$

Absolute deviations:

$$11.5,\ 5.5,\ 2.5,\ 1.5,\ 1.5,\ 1.5,\ 3.5,\ 5.5,\ 12.5,\ 24.5$$

Sum $= 70$, so

$$\text{M.D.}(\mathrm M) = \frac{70}{10} = 7$$

Find the mean deviation about the mean for the data in questions 5 and 6.

Question 5

$x_i$	$5$	$10$	$15$	$20$	$25$
$f_i$	$7$	$4$	$6$	$3$	$5$

Solution. $\mathrm N = \sum f_i = 25$, and

$$\sum f_ix_i = 35 + 40 + 90 + 60 + 125 = 350 \quad\Longrightarrow\quad \bar x = \frac{350}{25} = 14$$

Now $\left|x_i – 14\right|$ and $f_i\left|x_i – 14\right|$:

$x_i$	$\left\lvert x_i – 14\right\rvert$	$f_i$	$f_i\left\lvert x_i – 14\right\rvert$
$5$	$9$	$7$	$63$
$10$	$4$	$4$	$16$
$15$	$1$	$6$	$6$
$20$	$6$	$3$	$18$
$25$	$11$	$5$	$55$
		$25$	$158$

$$\text{M.D.}(\bar x) = \frac{158}{25} = 6.32$$

Question 6

$x_i$	$10$	$30$	$50$	$70$	$90$
$f_i$	$4$	$24$	$28$	$16$	$8$

Solution. $\mathrm N = 80$ and

$$\sum f_ix_i = 40 + 720 + 1400 + 1120 + 720 = 4000 \quad\Longrightarrow\quad \bar x = \frac{4000}{80} = 50$$

$x_i$	$\left\lvert x_i – 50\right\rvert$	$f_i$	product
$10$	$40$	$4$	$160$
$30$	$20$	$24$	$480$
$50$	$0$	$28$	$0$
$70$	$20$	$16$	$320$
$90$	$40$	$8$	$320$
		$80$	$1280$

$$\text{M.D.}(\bar x) = \frac{1280}{80} = 16$$

Find the mean deviation about the median for the data in questions 7 and 8.

Question 7

$x_i$	$5$	$7$	$9$	$10$	$12$	$15$
$f_i$	$8$	$6$	$2$	$2$	$2$	$6$

Solution. $\mathrm N = 26$, so $\tfrac{\mathrm N}{2} = 13$. Building the cumulative frequencies:

$$8,\ 14,\ 16,\ 18,\ 20,\ 26$$

The cumulative frequency first reaches or exceeds $13$ at $x = 7$, so $\mathrm M = 7$.

$x_i$	$\left\lvert x_i – 7\right\rvert$	$f_i$	product
$5$	$2$	$8$	$16$
$7$	$0$	$6$	$0$
$9$	$2$	$2$	$4$
$10$	$3$	$2$	$6$
$12$	$5$	$2$	$10$
$15$	$8$	$6$	$48$
		$26$	$84$

$$\text{M.D.}(\mathrm M) = \frac{84}{26} = \frac{42}{13} \approx 3.23$$

Question 8

$x_i$	$15$	$21$	$27$	$30$	$35$
$f_i$	$3$	$5$	$6$	$7$	$8$

Solution. $\mathrm N = 29$, so $\tfrac{\mathrm N}{2} = 14.5$. Cumulative frequencies:

$$3,\ 8,\ 14,\ 21,\ 29$$

The first to reach $14.5$ is $21$, at $x = 30$, so $\mathrm M = 30$.

$x_i$	$\left\lvert x_i – 30\right\rvert$	$f_i$	product
$15$	$15$	$3$	$45$
$21$	$9$	$5$	$45$
$27$	$3$	$6$	$18$
$30$	$0$	$7$	$0$
$35$	$5$	$8$	$40$
		$29$	$148$

$$\text{M.D.}(\mathrm M) = \frac{148}{29} \approx 5.1$$

Find the mean deviation about the mean for the data in questions 9 and 10.

Question 9

Income per day (₹)	$0$–$100$	$100$–$200$	$200$–$300$	$300$–$400$	$400$–$500$	$500$–$600$	$600$–$700$	$700$–$800$
Number of persons	$4$	$8$	$9$	$10$	$7$	$5$	$4$	$3$

Solution. Replace each class by its midpoint:

$$50,\ 150,\ 250,\ 350,\ 450,\ 550,\ 650,\ 750$$

$\mathrm N = 50$, and

$$\sum f_ix_i = 200 + 1200 + 2250 + 3500 + 3150 + 2750 + 2600 + 2250 = 17900$$

$$\bar x = \frac{17900}{50} = 358$$

Now $f_i\left|x_i – 358\right|$:

$x_i$	$\left\lvert x_i – 358\right\rvert$	$f_i$	product
$50$	$308$	$4$	$1232$
$150$	$208$	$8$	$1664$
$250$	$108$	$9$	$972$
$350$	$8$	$10$	$80$
$450$	$92$	$7$	$644$
$550$	$192$	$5$	$960$
$650$	$292$	$4$	$1168$
$750$	$392$	$3$	$1176$
		$50$	$7896$

$$\text{M.D.}(\bar x) = \frac{7896}{50} = 157.92$$

Question 10

Height (cm)	$95$–$105$	$105$–$115$	$115$–$125$	$125$–$135$	$135$–$145$	$145$–$155$
Number of boys	$9$	$13$	$26$	$30$	$12$	$10$

Solution. Midpoints: $100, 110, 120, 130, 140, 150$; $\mathrm N = 100$.

$$\sum f_ix_i = 900 + 1430 + 3120 + 3900 + 1680 + 1500 = 12530$$

$$\bar x = \frac{12530}{100} = 125.3$$

$x_i$	$\left\lvert x_i – 125.3\right\rvert$	$f_i$	product
$100$	$25.3$	$9$	$227.7$
$110$	$15.3$	$13$	$198.9$
$120$	$5.3$	$26$	$137.8$
$130$	$4.7$	$30$	$141.0$
$140$	$14.7$	$12$	$176.4$
$150$	$24.7$	$10$	$247.0$
		$100$	$1128.8$

$$\text{M.D.}(\bar x) = \frac{1128.8}{100} = 11.288 \approx 11.28$$

Question 11

Find the mean deviation about the median for the following data:

Marks	$0$–$10$	$10$–$20$	$20$–$30$	$30$–$40$	$40$–$50$	$50$–$60$
Number of girls	$6$	$8$	$14$	$16$	$4$	$2$

Solution. $\mathrm N = 50$, so $\tfrac{\mathrm N}{2} = 25$. Cumulative frequencies:

$$6,\ 14,\ 28,\ 44,\ 48,\ 50$$

The first to reach $25$ is $28$, so the median class is $20$–$30$, with $l = 20$, $\mathrm C = 14$, $f = 14$, $h = 10$:

$$\mathrm M = 20 + \frac{25 – 14}{14} \times 10 = 20 + \frac{110}{14} = 20 + \frac{55}{7} = \frac{195}{7} \approx 27.86$$

Now the deviations of the midpoints from $\mathrm M$:

$x_i$	$f_i$	$\left\lvert x_i – \mathrm M\right\rvert$	product
$5$	$6$	$22.857$	$137.14$
$15$	$8$	$12.857$	$102.86$
$25$	$14$	$2.857$	$40.00$
$35$	$16$	$7.143$	$114.29$
$45$	$4$	$17.143$	$68.57$
$55$	$2$	$27.143$	$54.29$
	$50$		$517.14$

$$\text{M.D.}(\mathrm M) = \frac{517.14}{50} = \frac{362}{35} \approx 10.34$$

Question 12

Calculate the mean deviation about the median age for the age distribution of $100$ persons given below:

Age (years)	$16$–$20$	$21$–$25$	$26$–$30$	$31$–$35$	$36$–$40$	$41$–$45$	$46$–$50$	$51$–$55$
Number	$5$	$6$	$12$	$14$	$26$	$12$	$16$	$9$

Solution. The classes are discontinuous — $16$–$20$ then $21$–$25$, with a gap. Convert them to continuous classes first, as the textbook’s hint directs, by subtracting $0.5$ from each lower limit and adding $0.5$ to each upper:

$$15.5\text{–}20.5,\quad 20.5\text{–}25.5,\quad 25.5\text{–}30.5,\quad \ldots,\quad 50.5\text{–}55.5$$

The midpoints are unchanged — $18, 23, 28, 33, 38, 43, 48, 53$ — but the class boundaries now abut, which the median formula requires.

$\mathrm N = 100$, so $\tfrac{\mathrm N}{2} = 50$. Cumulative frequencies:

$$5,\ 11,\ 23,\ 37,\ 63,\ 75,\ 91,\ 100$$

The first to reach $50$ is $63$, so the median class is $35.5$–$40.5$, with $l = 35.5$, $\mathrm C = 37$, $f = 26$, $h = 5$:

$$\mathrm M = 35.5 + \frac{50 – 37}{26} \times 5 = 35.5 + \frac{65}{26} = 35.5 + 2.5 = 38$$

$x_i$	$f_i$	$\left\lvert x_i – 38\right\rvert$	product
$18$	$5$	$20$	$100$
$23$	$6$	$15$	$90$
$28$	$12$	$10$	$120$
$33$	$14$	$5$	$70$
$38$	$26$	$0$	$0$
$43$	$12$	$5$	$60$
$48$	$16$	$10$	$160$
$53$	$9$	$15$	$135$
	$100$		$735$

$$\text{M.D.}(\mathrm M) = \frac{735}{100} = 7.35$$

Common mistakes

• Dropping the moduli. $\sum(x_i – \bar x) = 0$ always, so a mean deviation computed without absolute values comes out as zero. Take the modulus of each deviation before multiplying by $f_i$.
• Questions 3 and 4, not sorting first. The median is the middle value of the ordered data. Reading the middle of the list as given is the commonest error in these two.
• Questions 7 and 8, interpolating the median. For a discrete frequency distribution the median is one of the listed $x_i$ — the one where the cumulative frequency first reaches $\tfrac{\mathrm N}{2}$ — not a value between them.
• Questions 9 to 12, using class limits instead of midpoints. Grouped data is represented by the midpoint of each class throughout.
• Question 11, taking the wrong median class. It is the class where the cumulative frequency first reaches $\tfrac{\mathrm N}{2}$, and $\mathrm C$ is the cumulative frequency of the class before it, not including it.
• Question 12, using the classes as printed. $16$–$20$ and $21$–$25$ leave a gap, so the median formula’s $l$ and $h$ would be wrong. The $\pm0.5$ correction closes the gaps without moving any midpoint.
• Rounding too early. In questions 10 and 11 the intermediate deviations are not whole numbers; rounding them before summing shifts the final answer in the second decimal place.

Practise next

• Exercise 13.2 — variance and standard deviation, which square the deviations instead of taking their modulus and so are open to further algebra.
• Miscellaneous Exercise on Chapter 13 — problems where the mean and variance are given and the observations must be recovered.